Question

If the angle between the tangents drawn to the parabola $y^2=4x$ from the points on the line $4x-y=0$ is $\frac{\pi}{3}$, then the sum of the abscissae of all such points is

• A. 5/3
• B. 4/7
• C. 2/5
• D. 10/13

Hint:

The locus of the intersection of tangents to the parabola $y^2=4ax$ that include a constant angle $\theta$ is a hyperbola given by $y^2-4ax = (x+a)^2 \tan^2\theta$. This is known as the director curve (not to be confused with the director circle).

Answer 1

The Correct Option is D

Step 1: Use the locus of points for a fixed angle between tangents.
The locus of a point $(h,k)$ from which the two tangents drawn to the parabola $y^2=4ax$ make an angle $\theta$ is given by the equation: \[ (y^2-4ax) = (x+a)^2 \tan^2\theta. \] This equation describes a hyperbola.

Step 2: Apply the given parameters to find the locus equation.
For the parabola $y^2=4x$, we have $a=1$. The angle is $\theta=\pi/3$, so $\tan\theta = \sqrt{3}$ and $\tan^2\theta = 3$. Substituting these into the locus formula (using $(x,y)$ instead of $(h,k)$): \[ y^2-4x = (x+1)^2 (3). \]

Step 3: Find the intersection of this locus with the given line.
The points lie on the line $4x-y=0$, which means $y=4x$. We substitute $y=4x$ into the locus equation to find the coordinates of these points. \[ (4x)^2 - 4x = 3(x+1)^2. \] \[ 16x^2 - 4x = 3(x^2+2x+1). \] \[ 16x^2 - 4x = 3x^2+6x+3. \]

Step 4: Solve the resulting quadratic equation.
Rearrange the terms to form a standard quadratic equation: \[ (16-3)x^2 + (-4-6)x - 3 = 0. \] \[ 13x^2 - 10x - 3 = 0. \] This equation gives the abscissae of the points that satisfy the conditions.

Step 5: Find the sum of the abscissae.
Let the roots of the quadratic equation $13x^2 - 10x - 3 = 0$ be $x_1$ and $x_2$. These are the abscissae of the required points. The sum of the roots of a quadratic equation $Ax^2+Bx+C=0$ is given by $-B/A$. \[ \text{Sum of abscissae} = x_1 + x_2 = -\frac{-10}{13} = \frac{10}{13}. \] This matches option (D). Note: The keyed answer in the provided image (A) is incorrect. The derivation consistently yields (D).